If $f$ is bounded and left-continuous, can $f$ be nowhere continuous? If a function $f$ is bounded on $[a,b]$ and the left-hand limit exists at each point of $(a,b]$, can the function be nowhere continuous on $[a,b]$? 
 A: This will be an elementary proof. First consider the definition of left continuity:
$\forall x \in (a,b]$ and $\epsilon > 0$, $\exists \delta > 0$ such that if $x_0 \in (x-\delta,x)$, $|f(x_0)-f(x)| < \epsilon$.
It's possible to simply generate a sequence of increasingly closer points with the intent of proving right continuity. Take an $x_0 \in (a,b)$, and define three sequences for $i \ge 0$:
$$\epsilon_i = 2^{-i}$$
$$\Delta_0 = \delta_0, \Delta_{i+1} = \min(\delta_{i+1}, \frac{1}{2}\Delta_{i})$$
$$x_{i+1}=x_i-\Delta_k$$
Where $\delta_i$ is a $\delta$ given in response to $\epsilon_i$ at $x_i$ in the definition of left continuity.
Since the sequence $\{\Delta\}$ is at least as small as a geometric series, we can set
$$A = \lim_{k \to \infty} x_k$$
From here we can put bounds on $A-x_i$. Geometric series tells us that the sum of decreasing powers of $2$ is twice the first number. Thus, we have $x_i-A\le 2\Delta_i\le\Delta_{i-1}$.
Now, we simply note that taking $n \in (A,A+\Delta_{i-1})$ allows us use the definition of $\{\Delta\}$ , giving $|f(n)-f(A)|<\epsilon_{i-1}=2^{-(i-1)}=2^{1-i}$. The converse of this implies right continuity at A. $\blacksquare$
Using more advanced methods, it can be proven that there exist only a countable set of counterexamples in $(a,b)$.
A: For $x_0\in[a,b)$ define 
$$\omega_f^+(x_0)=\lim_{\epsilon\to 0}\omega_f^+\big([x_0,x_0+\epsilon)\big)\;,$$
where
$$\omega_f^+\big([x_0,x_0+\epsilon)\big)=\sup_{x\in[x_0,x_0+\epsilon)}f(x)-\inf_{x\in[x_0,x_0+\epsilon)}f(x)\;;$$
$\omega_f^+(x_0)$ is the oscillation of $f$ from the right at $x_0$.
Suppose that $f$ is nowhere continuous on $[a,b]$; then $\omega_f^+(x)>0$ for each $x\in[a,b)$. For $k\in\Bbb N$ let 
$$A_k=\left\{x\in[a,b):\omega_f^+(x)\ge 2^{-k}\right\}\;;$$
clearly $[a,b)=\bigcup_{k\in\Bbb N}A_k$. By the Baire category theorem some $A_k$ is dense in $[a,b)$. 
Let $\langle x_n:n\in\Bbb N\rangle$ be a strictly increasing sequence in $A_k$ with limit $x\in[a,b)$. For $n\in\Bbb N$ let
$$y_n=\sup_{u\in[x_n,x]}f(u)\qquad\text{and}\qquad z_n=\inf_{u\in[x_n,x]}f(u)\;;$$
then by hypothesis
$$\lim_{n\to\infty}y_n=f(x)=\lim_{n\to\infty}z_n\;,$$
so $\lim\limits_{n\to\infty}(y_n-z_n)=0$, while on the other hand
$$y_n-z_n\ge\omega_f^+(x_n)\ge 2^{-k}$$
for each $n\in\Bbb N$, which is absurd. Thus, $f$ cannot be nowhere continuous on $[a,b]$. Indeed, it’s clear from this argument that $f$ cannot be nowhere continuous on any non-degenerate interval, so $f$ must be continuous on a dense set of points.
A: You might wish to know as well what the historical answer to this question is.  This will take you back more than a century.  I am quoting from our paper on the Youngs:

Bruckner, Andrew M. ;  Thomson, Brian S.  Real variable
  contributions of G. C. Young and W. H. Young. Expo. Math.  19 
  (2001),  no. 4, 337--358.

"In [W 1908i] 

[W 1908i] On the distinction of right and left at points of
  discontinuity. Quarterly Journal of Pure and Applied Math. 39, 67-83.

is given the following theorem: if $f$ is an
arbitrary function of a real variable then for all values of $x$, excepting perhaps for a
countable set,
$$\limsup_{h\to 0+} f(x+h) = \limsup_{h\to 0+} f(x-h)$$
and 
$$\liminf_{h\to 0+} f(x+h) = \liminf_{h\to 0+} f(x-h).$$
Because the theorem was announced at the meeting of the British Association at Leicester
in 1907 they used to refer to this as the "Leicester theorem." The next year at the Rome
congress of 1908 this was improved to the statement that, again for all values of $x$, excepting
perhaps for a countable set all of the left and right limit numbers are identical. Stated in
more modern language this theorem (naturally called the "Rome theorem") asserts that
at all but countably many points the right and left cluster sets of an arbitrary function $f$
are identical."
Thus the question that you posed was very close to that posed by the Youngs long ago.  They embarked on quite a long research program of determining what kind of left/right asymmetry is possible.  Not only for limits and continuity (as here) but also for Dini derivatives.  Grace won a prize for her well-known theorem on the asymmetry for the Dini derivatives.  As we point out in the article, if you find a paper written by W. H. Young (the husband) assume that Grace had a lion share in the thinking and probably full responsibility for writing it up.   
